JEE Main 2023 (12 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Two dice $A$ and $B$ are rolled. Let the numbers obtained on $A$ and $B$ be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to

Select Answer:

Visualized Solution

The Two Dice

Two dice $A$ and $B$ are rolled.
The outcomes are denoted by $α$ and $β$ .

Variance of a Difference

We need to find the variance of the difference: $V a r (α - β)$ .
Using properties of variance: $V a r (a X + bY) = a^{2} V a r (X) + b^{2} V a r (Y)$ for independent $X, Y$ .
Here, $a = 1$ and $b = - 1$ .

Simplifying the Variance Expression

Substitute the coefficients: $V a r (α - β) = (1)^{2} V a r (α) + (- 1)^{2} V a r (β)$ .
This simplifies to: $V a r (α - β) = V a r (α) + V a r (β)$ .

Variance of a Uniform Distribution

The outcomes of a die roll follow a discrete uniform distribution from $1$ to $n$ .
The standard formula for variance is: $V a r (X) = \frac{n ^{2} - 1}{12}$ .

Substituting $n = 6$

For a standard die, the number of faces is $n = 6$ .
Substitute $n = 6$ into the variance formula: $V a r (α) = \frac{6 ^{2} - 1}{12}$ .

Calculating $V a r (α)$

$6^{2} = 36$ .
$36 - 1 = 35$ .
Therefore, $V a r (α) = \frac{35}{12}$ .
Since die B is identical, $V a r (β) = \frac{35}{12}$ .

Total Variance Setup

We know $V a r (α - β) = V a r (α) + V a r (β)$ .
Substitute the calculated values: $V a r (α - β) = \frac{35}{12} + \frac{35}{12}$ .

Simplifying the Total Variance

Add the fractions: $\frac{35 + 35}{12} = \frac{70}{12}$ .
Simplify by dividing numerator and denominator by $2$ .
$V a r (α - β) = \frac{35}{6}$ .

Identifying $p$ and $q$

The problem states $V a r (α - β) = \frac{p}{q}$ .
We found $V a r (α - β) = \frac{35}{6}$ .
Check if $35$ and $6$ are co-prime (their greatest common divisor is $1$ ).
Since $g c d (35, 6) = 1$ , we have $p = 35$ and $q = 6$ .

Finding Divisors of $p$

We need the positive divisors of $p = 35$ .
The divisors are the numbers that perfectly divide $35$ .
These are: $1, 5, 7, and 35$ .

Summing the Divisors

Sum $= 1 + 5 + 7 + 35$ .
$1 + 5 = 6$ .
$6 + 7 = 13$ .
$13 + 35 = 48$ .
The final answer is 48.

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The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

When rolling two independent dice with outcomes

α

and

β

, we are dealing with independent random variables. The variance of the difference between two independent variables is the sum of their individual variances.

This is because uncertainty is additive in nature. Mathematically, this is expressed as:

V a r (α - β) = V a r (α) + V a r (β)

Calculating the Variance of a Die

A standard six-sided die follows a discrete uniform distribution. The variance of the first

n

integers is given by the formula:

V a r (X) = \frac{n ^{2} - 1}{12}

For a standard die where

n = 6

, the variance is:

V a r (X) = \frac{6 ^{2} - 1}{12} = \frac{35}{12}

To verify this, we calculate the mean

E [X]

and the expected value of the square

E [X^{2}]

E [X] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5

E [X^{2}] = \frac{1 ^{2} + 2 ^{2} + 3 ^{2} + 4 ^{2} + 5 ^{2} + 6 ^{2}}{6} = \frac{91}{6}

Using the identity

V a r (X) = E [X^{2}] - (E [X])^{2}

, we confirm:

V a r (X) = \frac{91}{6} - (3.5)^{2} = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}

The Final Calculation

Since

α

and

β

are independent, the variance of their difference is:

V a r (α - β) = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6}

The problem asks us to consider the value derived from this variance, specifically focusing on the numerator

35

. We must find the sum of the divisors of

35

The divisors of

35

are

1, 5, 7,

and

35

. Summing these values gives:

1 + 5 + 7 + 35 = 48

The final result of this calculation is 48.

Common Traps & FAQs

Wait, why do we add the variances? If we are calculating the variance of

α - β

, shouldn't we subtract them?

This is the most common trap! Remember, variance measures the 'spread' or 'uncertainty' of a variable. When you combine two independent sources of uncertainty, the total uncertainty increases, regardless of whether you add or subtract the variables themselves. Mathematically,

V a r (a X + bY) = a^{2} V a r (X) + b^{2} V a r (Y)

. Since the coefficient of

β

- 1

, its square is

+ 1

. Thus, the variances always add.

How do I derive the variance formula

\frac{n ^{2} - 1}{12}

if I forget it in the exam?

Great question! Don't rely on rote memorization. The variance is

E [X^{2}] - (E [X])^{2}

. For a die,

E [X] = \frac{n + 1}{2}

and

E [X^{2}] = \frac{( n + 1 ) ( 2 n + 1 )}{6}

. If you plug these into the variance formula, you will derive

\frac{n ^{2} - 1}{12}

. Knowing the derivation gives you confidence!

What if the dice were not independent? Would the variance still be the sum?

No! If the dice were dependent, we would have to include the covariance term:

V a r (α - β) = V a r (α) + V a r (β) - 2 C o v (α, β)

. Since the problem specifies two dice being rolled (implying independence), the covariance is zero, which is why we only sum the variances.

Animated Solution for Mathematics - Probability: Two dice $A$ and $B$ are rolled. Let the numbers obtained on $A$ and $B$ be $α$ and $β$ respectively. If the variance of $α - β$ is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then the sum of the positive divisors of $p$ is equal to

Select Answer:

Visualized Solution

The Two Dice

Variance of a Difference

Simplifying the Variance Expression

Variance of a Uniform Distribution

Substituting $n = 6$

Calculating $V a r (α)$

Total Variance Setup

Simplifying the Total Variance

Identifying $p$ and $q$

Finding Divisors of $p$

Summing the Divisors

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Calculating the Variance of a Die

The Final Calculation

Similar Questions

A six faced die is biased such that $3 \times P (a prime number) = 6 \times P (a composite number) = 2 \times P (1)$ . Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is :

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6∣ X > 3)$ . Then $\frac{b + c}{a}$ is equal to

A person throws two fair dice. He wins Rs. 15 for throwing a doublet, wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is:

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

Select Answer:

Visualized Solution

The Two Dice

Variance of a Difference

Simplifying the Variance Expression

Variance of a Uniform Distribution

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=6

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Var(α)

Total Variance Setup

Simplifying the Total Variance

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Finding Divisors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Summing the Divisors

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

Calculating the Variance of a Die

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a=P(X=3),b=P(X≥3) and c=P(X≥6∣X>3). Then ab+c​ is equal to

A person throws two fair dice. He wins Rs. 15 for throwing a doublet, wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is:

Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :

Substituting $n = 6$

Calculating $V a r (α)$

Identifying $p$ and $q$

Finding Divisors of $p$

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3), b = P (X \geq 3)$ and $c = P (X \geq 6∣ X > 3)$ . Then $\frac{b + c}{a}$ is equal to

A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is: